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Water is being boiled in an open kettle that has a 0.500-cm-thick circular aluminum bottle with a radius of 12.0cm. If the water boils away at a rate

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asked 2021-03-27
Water is being boiled in an open kettle that has a 0.500-cm-thick circular aluminum bottle with a radius of 12.0cm. If the water boils away at a rate of 0.500 kg/min,what is the temperature of the lower surface of the bottom of the kettle? Assume that the top surface of the bottom of the kettle is at \(\displaystyle{100}^{\circ}\) C.

Answers (1)

2021-03-29
Water is being boiled in an open kettlethat has a (L = 0.500-cm) -thick circular aluminum bottle with a radius of (r = 12.0cm).
If the water boils away at a rate of
\(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}={0.500}\ {k}\frac{{g}}{\min}={8.33}\cdot{10}^{{-{3}}}\ {k}\frac{{g}}{{s}}\)
Let the temperature of the lower surface of the bottom of thekettle be \(\displaystyle{{t}_{{2}}^{\circ}}\) C
The temperature of thetop surface of the bottom of the kettle (\(\displaystyle{t}_{{1}}={100}^{\circ}{C}\))
Since the heat given by the aluminium bottle = heattaken by boiled water
\(\displaystyle{k}{A}{\left({t}_{{2}}-{t}_{{1}}\right)}\cdot{\frac{{{\left.{d}{t}\right.}}}{{{L}}}}{)}={d}{m}\cdot{L}_{{s}}\)
Where
\(\displaystyle{K}={238}\ \frac{{J}}{{s}}\)
\(\displaystyle{A}={p}{r}^{{2}}\)
dt=time taken for the mass dm
Latent Heat of steam (\(\displaystyle{L}_{{s}}\))=2256000 J/kg
Ahain from above equation
\(\displaystyle{k}{A}{\frac{{{t}_{{2}}-{t}_{{1}}}}{{{d}}}}={\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}\cdot{L}_{{s}}\)
Now calculate the value of \(\displaystyle{t}_{{2}}\) from above
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